An identity or a factorisation is true for every value of the letter. Writing a number in the form of one side lets you work with the other side, which is often much easier. Evaluating without a calculator Write the numbers so that they match one side of an identity, then use the other side. **Expression** | **Identity** | **Working** | **Value** 972-9 | a2-b2 | 100 x 94 | 9400 532-2(53)(3)+9 | (a-b)2 | 502 | 2500 3(26.5)2-3(13.5)2 | 3(a2-b2) | 3 x 40 x 13 | 1560 In the first row a=97 and b=3, so 972-9=(97+3)(97-3)=100 x 94. In the second, a=53 and b=3, so the expression is (53-3)2. In the third, take out the common factor 3 first; then a=26.5 and b=13.5, so 3(26.5+13.5)(26.5-13.5)=3 x 40 x 13. Why: 972-9 needs 972=9409 worked out by hand; the identity turns it into 100 x 94, which is easy. Finding factors of a number A factorisation turns a number into a product. For example, x2-1=(x+1)(x-1) for every x. With x=30, the left side is 899 and the right side is 31 x 29, so 899=31 x 29 and 899 is not prime. Method - Factorise the algebraic expression first. - Find the value of the letter that makes the expression equal to the given number (see below). - Substitute that value into every factor. - Check by multiplying the factors. Finding the value of the letter You do not need to solve an equation. Use one of these. - **Difference of two squares:** add the square to the number. For x2-49=1551, x2=1600, so x=40 and 1551=(40+7)(40-7)=47 x 33. - **Other quadratics:** try small whole numbers and watch the value grow. For 6x2+5x+1=176: x=4 gives 117, x=5 gives 176. So x=5, and 6x2+5x+1=(3x+1)(2x+1) gives 176=16 x 11. - **A number such as 273:** look at its digits. 273=2 x 100+7 x 10+3 matches 2x2+7x+3 with x=10. Example Factorise 2x2+7x+3. Hence find two factors of 273 other than 1 and 273. 2x2+7x+3 = 2x2+6x+x+3 = 2x(x+3)+(x+3) = (2x+1)(x+3) When x=10, 2x2+7x+3=200+70+3=273, and (2x+1)(x+3)=21 x 13. So 21 and 13 are factors of 273. Check: 21 x 13=273. A common error is to substitute into only one factor, or to choose a value of x that does not give the required number.

Using identities and factorisation with numbers

Secondary 2

An identity or a factorisation is true for every value of the letter. Writing a number in the form of one side lets you work with the other side, which is often much easier.

Evaluating without a calculator

Write the numbers so that they match one side of an identity, then use the other side.

Expression	Identity	Working	Value
$97^2-9$	$a^2-b^2$	$100\times94$	$9400$
$53^2-2(53)(3)+9$	$(a-b)^2$	$50^2$	$2500$
$3(26.5)^2-3(13.5)^2$	$3(a^2-b^2)$	$3\times40\times13$	$1560$

In the first row $a=97$ and $b=3$, so $97^2-9=(97+3)(97-3)=100\times94$. In the second, $a=53$ and $b=3$, so the expression is $(53-3)^2$. In the third, take out the common factor $3$ first; then $a=26.5$ and $b=13.5$, so $3(26.5+13.5)(26.5-13.5)=3\times40\times13$.

Why: $97^2-9$ needs $97^2=9409$ worked out by hand; the identity turns it into $100\times94$, which is easy.

Finding factors of a number

A factorisation turns a number into a product. For example, $x^2-1=(x+1)(x-1)$ for every $x$. With $x=30$, the left side is $899$ and the right side is $31\times29$, so $899=31\times29$ and $899$ is not prime.

Method

Factorise the algebraic expression first.
Find the value of the letter that makes the expression equal to the given number (see below).
Substitute that value into every factor.
Check by multiplying the factors.

Finding the value of the letter

You do not need to solve an equation. Use one of these.

Difference of two squares: add the square to the number. For $x^2-49=1551$, $x^2=1600$, so $x=40$ and $1551=(40+7)(40-7)=47\times33$.
Other quadratics: try small whole numbers and watch the value grow. For $6x^2+5x+1=176$: $x=4$ gives $117$, $x=5$ gives $176$. So $x=5$, and $6x^2+5x+1=(3x+1)(2x+1)$ gives $176=16\times11$.
A number such as $273$: look at its digits. $273=2\times100+7\times10+3$ matches $2x^2+7x+3$ with $x=10$.

Example

Factorise $2x^2+7x+3$. Hence find two factors of $273$ other than $1$ and $273$.

$\begin{aligned} 2x^2+7x+3 &= 2x^2+6x+x+3 \\ &= 2x(x+3)+(x+3) \\ &= (2x+1)(x+3) \end{aligned}$

When $x=10$, $2x^2+7x+3=200+70+3=273$, and $(2x+1)(x+3)=21\times13$. So $21$ and $13$ are factors of $273$.

Check: $21\times13=273$. A common error is to substitute into only one factor, or to choose a value of $x$ that does not give the required number.

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Expression	Identity	Working	Value
\(97^2-9\)	\(a^2-b^2\)	\(100\times94\)	\(9400\)
\(53^2-2(53)(3)+9\)	\((a-b)^2\)	\(50^2\)	\(2500\)
\(3(26.5)^2-3(13.5)^2\)	\(3(a^2-b^2)\)	\(3\times40\times13\)	\(1560\)